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2 years ago

HELP ME PLEASE!!

Mathematics

1 answer:

DiKsa [7]2 years ago

5 0

C=A+B
C₄₁= row=4; column=1
C₄₁=A₄₁+B₄₁
=2(7)-3(4)
=14-12
=2

Answer: C₄₁=2

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Please help x-3=x^(2)-10x+25

schepotkina [342]

Answer:

x=7,4

Step-by-step explanation:

8 0

2 years ago

Help!!!!!!!!!! <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B500%7D%20%7D%7B%5Csqrt%5B%5D%7B5%7D%20%7D" id="TexFormu

DochEvi [55]

Answer:

Step-by-step explanation:

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2 years ago

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What percent of a data set is represented by Q1

Ludmilka [50]

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3 years ago

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A car travels 3 times around a traffic circle whose radius is 80 feet. What is the distance the car will travel? Use 3.14 for π

Harman [31]

Answer : Distance will be 1507.2 feet .

Explanation :

Since we have given that

Radius of circle = 80 feet

So,

Circumference of circle is given by

$2\pi r=2\times 3.14\times 80=502.4 \text{ feet}$

Since , a car travels 3 times around a traffic circle.

So,

$\text{ Distance covered by the car will travel}= 3\times 502.4= 1507.2 \text{ feet }$

So, Distance will be 1507.2 feet .

5 0

3 years ago

If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi

Finger [1]

ANSWER:

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

SOLUTION:

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = $\mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2}$ --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

$\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16$

$34 \times 1+\frac{1}{2} \times a \times 1^{2}=18$

$34+\frac{a}{2}=18$

$\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}$

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0

3 years ago